Question Number 10043 by konen last updated on 21/Jan/17 $$\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}^{\mathrm{2}} −\mathrm{9}} =\mathrm{1}\:\:\Rightarrow\Sigma\mathrm{x}=? \\ $$ Answered by mrW1 last updated on 22/Jan/17 $${if}\:{a}^{{b}} =\mathrm{1},\:{then} \\ $$$${a}=\mathrm{1}\:{or}…
Question Number 10041 by prakash jain last updated on 21/Jan/17 Answered by mrW1 last updated on 22/Jan/17 $${they}\:{form}\:{a}\:{lozenge},\:{but}\:{not}\:{nacessarily} \\ $$$${a}\:{square}. \\ $$ Commented by mrW1…
Question Number 75577 by Tony Lin last updated on 13/Dec/19 $$\int_{−\infty} ^{\infty} {xdx}\:{is}\:{divergent}\:{or}\:{convergent}? \\ $$ Answered by MJS last updated on 13/Dec/19 $$\underset{−\infty} {\overset{+\infty} {\int}}{xdx}=\underset{{r}\rightarrow\infty}…
Question Number 10040 by FilupSmith last updated on 21/Jan/17 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\lfloor\mathrm{log}_{\mathrm{10}} \left({x}\right)+\mathrm{1}\rfloor\:\:\:\mathrm{give}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{digits}\:\mathrm{for}\:{x}\in\mathbb{Z} \\ $$ Answered by mrW1 last updated on 21/Jan/17 $${let}\:{x}\:{be}\:{an}\:{integer}\:{with}\:{n}\:{digits},\:{then}…
Question Number 10039 by Gaurav3651 last updated on 21/Jan/17 $$ \\ $$$${let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{twice}\:{differentiable} \\ $$$${functions}\:{on}\:\left[\mathrm{0},\mathrm{2}\right]\:{satisfying} \\ $$$${f}''\left({x}\right)={g}''\left({x}\right),\:{f}'\left(\mathrm{1}\right)=\mathrm{4},\:{g}'\left(\mathrm{1}\right)=\mathrm{6}, \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{3}\:{and}\:{g}\left(\mathrm{2}\right)=\mathrm{9}.\:{Then}\:{what}\:{is} \\ $$$${f}\left({x}\right)−{g}\left({x}\right)\:{at}\:{x}=\mathrm{4}\:{equal}\:{to}? \\ $$ Commented by prakash…
Question Number 141111 by jlewis last updated on 15/May/21 $$\mathrm{A}\:\mathrm{gap}\:\mathrm{must}\:\mathrm{be}\:\mathrm{left}\:\mathrm{between}\:\mathrm{steel}\: \\ $$$$\mathrm{rails}\:\mathrm{to}\:\mathrm{allow}\:\mathrm{for}\:\mathrm{thermal}\:\mathrm{expansion}. \\ $$$$\:\mathrm{How}\:\mathrm{large}\:\mathrm{a}\:\mathrm{gap}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{maximum}\:\mathrm{temperature}\:\mathrm{reached} \\ $$$$\:\mathrm{is}\:\mathrm{50}°\:\mathrm{above}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{at}\: \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{rails}\:\mathrm{were}\:\mathrm{laid}.\:\mathrm{The}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rail}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\alpha_{\mathrm{steal}} =\mathrm{12}×\mathrm{10}^{−\mathrm{6}}…
Question Number 141110 by Fikret last updated on 15/May/21 $${sin}^{\mathrm{2}} {x}+{sinx}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\:\:\:\:\:,\:\:\:{x}\in\left[\mathrm{0},\pi\right] \\ $$$${equation}\:\:{sum}\:\:{of}\:{roots}\:{in}\:{range}? \\ $$ Answered by MJS_new last updated on 15/May/21 $${s}^{\mathrm{2}} +{s}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{0}\wedge−\mathrm{1}\leqslant{s}\leqslant\mathrm{1}\:\Rightarrow\:{s}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{57}}}{\mathrm{6}} \\…
Question Number 75570 by naka3546 last updated on 12/Dec/19 Commented by naka3546 last updated on 12/Dec/19 $$\frac{\left[\:{blue}\:\:{area}\:\right]}{\left[\:{square}\:\:{area}\:\right]}\:\:=\:\:? \\ $$ Answered by mr W last updated…
Question Number 10035 by Sri rama jeyam last updated on 21/Jan/17 $$\mathrm{is}\:\mathrm{cos}\left(\mathrm{7x}+\mathrm{i5y}\right)\:\mathrm{analytic}\:\mathrm{function}\:\mathrm{or}\:\mathrm{not} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141104 by Opredador last updated on 15/May/21 Answered by hknkrc46 last updated on 15/May/21 $$\left.\begin{matrix}{\sqrt{\mathrm{7}^{\sqrt{\mathrm{63}}} }\:=\:\sqrt{\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{7}}} }\:=\:\mathrm{7}^{\sqrt{\mathrm{7}}} \sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }}\\{\mathrm{7}^{\sqrt{\mathrm{7}}} \:=\:\boldsymbol{{u}}}\end{matrix}\right\}\:\frac{\boldsymbol{{u}}\sqrt{\boldsymbol{{u}}}\:−\:\sqrt{\boldsymbol{{u}}}}{\boldsymbol{{u}}\:−\:\mathrm{1}} \\ $$$$=\:\frac{\sqrt{\boldsymbol{{u}}}\left(\boldsymbol{{u}}\:−\:\mathrm{1}\right)}{\boldsymbol{{u}}\:−\:\mathrm{1}}\:=\:\sqrt{\boldsymbol{{u}}}\:=\:\sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }…